QJEEM23Ma6b0f: JEE Main 2023 Mathematics - Class 11 JEEM23Ma6b0f Solution: Let the number of elements in sets and be five and two…

First, let's determine the number of elements in the Cartesian product $A \times B$ . If set $A$ has 5 elements and set $B$ has 2 elements, then the number of elements in $A \times B$ is:

$|A \times B| = |A| \times |B| = 5 \times 2 = 10$

We need to find the number of subsets of $A \times B$ with at least 3 elements and at most 6 elements. The total number of elements in $A \times B$ is 10, so we must consider the subsets containing 3, 4, 5, and 6 elements.

The number of ways to choose $r$ elements from a set of 10 elements is given by the binomial coefficient:

$\binom{10}{r} = \frac{10!}{r!(10-r)!}$

We will find the sum of the binomial coefficients for $r = 3, 4, 5, 6$ :

First, $\binom{10}{3}$ :

$\binom{10}{3} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$

Next, $\binom{10}{4}$ :

$\binom{10}{4} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

Next, $\binom{10}{5}$ :

$\binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$

Next, $\binom{10}{6}$ :

$\binom{10}{6} = \frac{10!}{6!4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

Finally, we sum these values:

$\binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6} = 120 + 210 + 252 + 210 = 792$

Therefore, the number of subsets of $A \times B$ each having at least 3 and at most 6 elements is:

Option D - 792